دانلود اپلیکیشن «زوم»

فایل ویدیویی

ترجمه‌ی درس

متن انگلیسی درس

QC Questions & Integer Properties

Quantitative comparisons and integer properties. Of course, the GRE loves to ask about integer properties, and we have had a whole lesson, whole module on, of lessons on these. In this lesson, I’ll assume that you’re familiar with all those properties that discussed in the inter, Integer Property module way back at the beginning of the math content.

Integer properties include odd and even, divisibility, factors and multiples, greatest common factor and least common multiples, primes and prime factorization, consecutive integers, and division and quotient and remainders. All those are included in integer properties. Remember the important exceptions. All prime numbers are odd except for 2, and 2 of course is the smallest and the only even prime number.

All other even numbers can’t be prime because they’re divisible by 2. All numbers on the number line are positive or negative, except for zero, which is neither positive nor negative. All positive numbers, all positive integers are either prime or product of prime except for 1, which is not prime, and not divisible by any prime number. So, 1 is a big exception among the positive integers.

1 is a factor of every positive integer. Every positive integer is a factor of itself. Any of these could be good fodder for GRE quantitative comparison questions. Something may seem true if you forget an important exception. Recall that when a number, a dividend, N, is divided by a divisor, D, this produces an integer quotient Q and a remainder R.

For example, when 17 is divided by 5, 5 goes into 17 three times. The quotient is 3 with a remainder of 2. Recall that the remainder is always less than the, than the divisor, always. Recall that if the divisor is bigger than the dividend, in other words, if what we’re dividing by is bigger than the thing we’re dividing, then the quotient equals 0, and the remainder equals the original number.

For example, 137 divides into 5 zero times, with 5 left over. So there, 5 is the dividend, 0 is the quotient, and 5 is the remainder. So it’s possible for the remainder to equal the dividend. Finally, recall the rebuilding the dividend formula, that the dividend equals the divisor times the quotient plus the remainder.

These topics and integer properties lend themselves to an endless number of quantitative comparison possibilities. So obviously, we can’t exhaust all the possibilities. I’ll just show you a few questions here at how to approach them. Here’s the first one. Pause the video and then we’ll talk about this.

So you may remember, back in the integer property module, we talked a little about prime factorization. We talked about factorials, and we talked a little bit about the prime factorization of factorials. So let’s think about this, the exponent of 5 in that prime factorization. Well, how many factors of 5 would there be in 17 factorial?

Well, in the positive integers from 1 to 17, most of them don’t have a factor of 5 at all. But three of them do, 5, 10, and 15. Each one of them has exactly one factor of 5. So take them all together, there are three factors of 5 in 17 factorial. And so, quantity A equals 3.

5 would have an exponent of 3. So there are three factors of 5. There are more factors of 3 and there are many, many factors of 2. So, 2 and 3 have lots of factors. 5 has three factors. The prime factors with no multiples less than 17 would have exponents of 1.

So, let’s think about the prime numbers. The number 17 has the multiple 14, so 17 would have an exponent of 2 because two factors show up of that. Of course, as we’ve just said, 5 has a, has an exponent of 3. 2 and 3 have much, much larger exponents. They definitely don’t have an exponent of just 1.

So to have an exponent of 1, it would have to be a prime number that does not have a multiple that is smaller than 17. And so, the numbers 11, 13, and 17 itself, each satisfy this condition. Each would have a multiple, have an exponent of 1 because there’s only one factor of each one of those. And so, Quantity B also equals 3.

And so, the two quantities here are equal. Here’s another practice problem. Pause the video and then we’ll talk about this. This is a tricky one.

The number of multiples of 3 between 3 to the 31 and 3 to the 41, and the number of multiples of 7 between 7 to the 21 and 7 to the 31. Well gosh, how would we even begin this? Let’s think about this. Think about some smaller numbers. For example, how many multiples of 5 are between 30 and 125?

Well, 30 of course equals 5 times 6, and 125 equals 5 times 25. So really in a way, it’s like asking how many integers are between 6 and 25. Well, to find the integers between 6 and 25, we’d subtract them. And then we, we actually have to subtract 1 after that. So 25 minus 6 is 19. That would actually exclude 6, but count 25.

And we, we want the one between them, so we’d have to subtract 1, 2, exclude 25 also, so there are actually 18 numbers between 6 and 25. And that means there’d be 18 multiples of 5 between 30 and 125. So what we have to do is divide each number by 5, and then subtract them, and then subtract 1. Alternately, we could subtract the two numbers, and then divide.

That would be much the same thing. We could do 125 minus 30. Divide that by 5 and then subtract 1. So here we talk about, technically, here I explain, we have to subtract 1 because, again, 125 would be included.

Not counting either endpoint, either the 30 or the 125, there would be 18 multiples of 5 between these two numbers. All right, now that we’ve thought about that, let’s think about this problem. The number of multiples of 3 between 3 to the 31 and 3 to the 41. I’m gonna subtract them and divide by 3. Technically, there’d be a minus 1 from this fraction.

The, the other one would be 7 to the 31 minus 7 to the 21 divided by 7, again, minus 1. Since there’s a minus 1 from both of these fractions, I’m just gonna ignore them. It’s as if I added one to both quantities. That minus 1 is not really important. Okay.

So now we’ll divide, and in each term, the exponent goes down by 1. So really, what we wind up with is 3 to the 40th minus 3 to the 30th, and 7 to the 30th minus 7 to the 20th. Well, this is interesting. There’s no exponent law when we’re adding or subtracting two different powers, but we can factor things out.

In each case, we can factor out the smaller power. So in Quantity A, we’ll factor out the 3 to the 30th. That leaves us inside the parentheses with 3 to the 10th minus 1. In the bottom one, Quantity B, we’ll factor out 7 to the 20th. And that leaves us with 7 to the 10th minus 1. And those are the two things we are comparing.

Well, first of all, look at what’s in the parentheses. Of course, 7 to the 10 is gonna be bigger than 3 to the 10. That’s obvious. And so 7 to the 10 minus 1 is still gonna be bigger than 3 to the 10 minus 1. So, at least the part in parentheses, clearly Quantity B is bigger. Now look at what’s outside the parentheses.

3 to the 30th, I’m gonna rearrange the exponents. 30 is 3 times 10, so I’m gonna bring that 3 inside as its own exponent. And of course 3 to 3rd, that’s 27. So this is 27 to the power of 10. I’ll do that same trick with the other one. 7 to the 20th, that’s 7 squared to the 10th.

That’s 49 to the 10th. Well, notice 49 to the 10th has to be bigger than 27 to the 10th. So again, that piece of Quantity B is clearly bigger than Quantity A. So both pieces of Quantity B are bigger than the corresponding pieces of Quantity A. This means that Quantity B is definitively bigger.

Here’s a practice problem. Pause the video and then we’ll talk about this. So these are awfully big denominators to have here. First of all, we’re gonna find some, some common factors, and then multiply both quantities by common factors.

One common factor is clearly 25. So both of those values, 525 and 900, are clearly divisible by 25. Now, we don’t need a calculator for this. Just think about this. We know that 4 times 25 is 100. Multiply both sides of that equation by 9.

That means 36 times 25 is 900. Go back to 4 times 25 is 100. Multiply both sides by 5. That means 20 times 25 is 500. Well, that’s 500. Add one more factor of 25 to that.

If 20 25s make 500, then if we add 25, 21 25s will be 525. And so that expresses each denominator as 25 times something. So all we have to do is multiply both quantities by 25. We get much simpler fractions. Now, still those denominators are both divisible by 3.

Multiply both sides by 3. We get 4 over 7 equals 7 over 12. Now we have nice simple numbers. And remember, we’re always allowed to cross multiply. Everything is positive here, so we’re allowed to cross multiply. So, we get 4 times 12, which is 48.

7 times 7 is 49. 49 is bigger than 48, and Quantity B is bigger. Here’s a practice problem. Pause the video, then we’ll talk about this. Okay.

When 900 is divided by positive integer D, the remainder is 1. If N, some, some number larger than 5,000, then when N is divided by 23, the remainder is R. So clearly, the numbers here are quite unwieldy. They’re very large. And so, picking numbers is not necessarily gonna be helpful here.

So, a few things we’ll notice. First of all, we know that whatever that remainder R is, it has to be less than the divisor. And so, R has to be less than 23. We know that. All right, so now let’s think about that first division.

Whatever D is, it must be a factor of 899. That’s the only way we could divide it into 900 and get a remainder of 1. It has to go evenly into 899. So gee, 899, that could be a hard number to factor. Notice though that 899 is 900 minus 1, and that fits a Difference of Two Squares factoring pattern.

You may remember the lesson on advanced numerical factoring where we actually talked about using these algebraic patterns to factor numbers. So, 900 minus 1, that’s 30 squared minus 1 squared. So we can factor that into 30 minus 1 times 30 plus 1, which is 29 times 31. Very, very easy to factor it using that algebraic pattern. So now we know that 899 is the product of two prime numbers, 29 times 31.

So, the factors, the total factor list of 899 has to be 1, 29, 31, and 899. Those are the four factors of 899. Well, D obviously can’t be 1, because when we divide any integer by 1, there’s never a remainder. 1 goes evenly into every other positive integer.

So D can’t be 1. So D could be 29, or 31, or 899, all of which are bigger than 23, which is bigger than R. So this means that D, whatever it is, has to be bigger than R. And so the answer is A. Remember all the distinctions and exceptions in the Integer Properties.

Many integer properties questions will demand genuine mathematical thinking. So it’s very important to understand these properties very well.